Help with integral (pretty easy but my brain is stuck!)

[homestar]

1. The problem statement, all variables and given/known data

This is to find the arc length of r= \theta from 0 < \theta < 2pi

I ended up with \int\sqrt{\theta^2 + 1}

2. Relevant equations

\int\sqrt{\theta^2 + 1}


3. The attempt at a solution

I couldn't see a relevant way to do integration by parts so I went with trig substitution, subbing tan(t) for theta. I got lost at the point where I'm to integrate sec^3(t)dt.

Can anyone help? Am I on the right track?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

[rocomath]

Looks good to me.

\int\sec^{3}tdt

Let's start by breaking secant to the cube.

\int\sec t\sec^2 tdt

Which would you make u and dV?

[HallsofIvy]

1. The problem statement, all variables and given/known data

This is to find the arc length of r= \theta from 0 < \theta < 2pi

I ended up with \int\sqrt{\theta^2 + 1}

2. Relevant equations

\int\sqrt{\theta^2 + 1}
You actually mean \int\sqrt{\theta^2+ 1} d\theta don't you?


3. The attempt at a solution

I couldn't see a relevant way to do integration by parts so I went with trig substitution, subbing tan(t) for theta. I got lost at the point where I'm to integrate sec^3(t)dt.

Can anyone help? Am I on the right track?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data
Looks to me like a pretty standard integral. Let \theta= tan(x) so that d\theta= sec^2(x) dx and \sqrt{\theta^2+ 1}= \sqrt{tan^2(x)+ 1}= sec(x). Then
\int\sqrt{\theta^2+ 1}d\theta= \int sec^3(x)dx= \int \frac{1}{cos^3(x)}dx
= \int\frac{cox(x)}{cos^3(x)}dx= \int\frac{(cos(x)dx)}{(1- sin^2(x))^2}

and the substitution u= sin(x) converts to a rational integral:
[tex]\int \frac{1}{(1-u^2)^2}du[/itex]


2. Relevant equations



3. The attempt at a solution[/QUOTE]

[homestar]

ahhhh...thank you. it's relatively early in the semester and i may have partied too hard over winter break ;)